Time-dependent Schrödinger equation of the harmonic oscillator hello I have the time dependent Schrödinger equation of a harmonic oscillator $$i \hbar \frac{\partial}{\partial t} \Psi(q,t) = - \frac{\hbar^2}{2M} \frac{\partial^2}{\partial q^2} \Psi(q,t) + \frac{M \Omega^2}{2} q^2 \Psi(q,t)$$ and am supposed to solve it for the initial condition $$\Psi(q,0) = (\frac{A}{\pi})^{1/4} e^{-(A/2)q^2}.$$ Our prof advised us to find an Ansatz for which above differential equation transforms into a riccati differential equation and the only Ansatz I had was to try $$\Psi(q,t) = A e^{\lambda(t)q^2}$$ which then leads to the differential equation $$i \hbar \dot{\lambda}(t) = - \frac{\hbar^2}{Mq^2} \lambda(t) + \frac{2 \hbar^2}{M} \lambda^2(t) + \frac{M \Omega^2}{2}$$ which I still don't know how to solve. Are there any easier methods or does anybody know how to continue from there?
 A: One alternative way to go about it would be to use the already known static solutions of the harmonic oscillator $\phi_n(q)$. They are eigenfunctions, so they evolve in time by picking up a phase factor $e^{-i E_n t/\hbar}$, with $E_n = \hbar \Omega (n + 1/2)$. These functions are a complete basis of the Hilbert space, so you can especially decompose the initial conditions as
$$ \Psi_0(q) = \Psi(q,t=0) = \sum_{n\ge 0} \langle\phi_n |\Psi_0\rangle \phi_n(q).$$
You can thus obtain the solution as
$$\Psi(q,t) = e^{-iHt/\hbar}\Psi_0(q) = \sum_{n\ge 0} e^{-iE_nt/\hbar} \langle\phi_n |\Psi_0\rangle \phi_n(q).$$
This shifts the task from getting a new solution of the equation to just calculating $\langle \phi_n | \Psi_0\rangle$, which is feasible with Gaussian integrals and using a series expansion for the Hermite polynomials (which should converge).
There isn't much elegance in this solution. It's quite technical and maybe the detour over the Riccati equation yields a more useful result.
The ansatz you use leads to a Riccatti equation, but the anstaz has a flaw: it is not normalized. The norm is time-dependent,
$$ \langle \Psi(t) | \Psi(t)\rangle = |A|^2 \sqrt{\frac{\pi}{\lambda(t)}}.$$
This can not yield the correct solution because the Schrödinger equation is unitary, that is the norm of $\Psi$ is necessarily time-independent.
A: After experimenting and working out quite some algebra with Mathematica and using Hermite polynomials properties, I managed to analytically perform the infinite sum and get a relatively nice answer. This can be fairly time consuming so instead I present first here a simpler approach.
First, I am choosing to write
\begin{equation}
A = \alpha \frac{M \Omega}{\hbar} \, ,
\end{equation}
where $\alpha$ is a dimensionless positive constant. This does not represent any loss of generality.
Turns out, an ansatz that would actually fit the answer would be
\begin{equation}
\Psi(x,t) = \frac{A^{1/4}}{\pi^{1/4}} \frac{1}{h(t)^{1/2}} \exp\left[-\frac{1}{2} \frac{M \Omega}{\hbar} x^2 g(t) \right] \, ,
\end{equation}
together with the initial conditions
\begin{eqnarray}
g(0) &=& \alpha \, ,
\\
h(0) &=& 1 \, .
\end{eqnarray}
While in principle normalization necessarily relates $h$ and $g$, it comes as a benefit in this case to ignore this first and leave it as a consistency check for the end of the exercise.
Plugging this in the Schrödinger equation gives you a system of differential equations that can be decoupled and solved without too much effort. You should find something like this
\begin{equation}
\Psi(x,t) = \frac{A^{1/4}}{\pi^{1/4}} \frac{1}{\left[ \cos (t \Omega )+ i \alpha  \sin (t \Omega ) \right]^{1/2}} \exp\left[-\frac{1}{2} \frac{M \Omega}{\hbar} x^2 \frac{\alpha  \cos (t \Omega )+i \sin (t \Omega )}{\cos (t \Omega )+i \alpha  \sin (t \Omega )}\right] \, .
\end{equation}
Other options:
In case there is interest in finding this result through the sum over the eigenstates, I present here the short version of how to do so. We have
\begin{eqnarray}
|\Psi(t)\rangle &=& \exp\left[-i \frac{t H}{\hbar}\right] |\Psi(0)\rangle
\\
&=& \exp\left[-i \frac{t H}{\hbar}\right] \sum_{n=0}^\infty | n \rangle \langle n |\Psi(0)\rangle
\\
&=& \mathrm{e}^{- \frac{i t \Omega}{2}} \sum_{n=0}^\infty \left( \mathrm{e}^{-i t \Omega} \right)^n | n \rangle \langle n |\Psi(0)\rangle \, .
\end{eqnarray}
The wave function can then be found with
\begin{eqnarray}
\Psi(x,t) &=& \langle x |\Psi(t)\rangle
\\
&=& \mathrm{e}^{- \frac{i t \Omega}{2}} \sum_{n=0}^\infty \left( \mathrm{e}^{-i t \Omega} \right)^n \langle x | n \rangle \langle n |\Psi(0)\rangle
\\
&=& \mathrm{e}^{- \frac{i t \Omega}{2}} \int_{- \infty}^\infty \mathrm{d} x' \sum_{n=0}^\infty \left( \mathrm{e}^{-i t \Omega} \right)^n \langle x | n \rangle \langle n | x' \rangle \langle x' |\Psi(0)\rangle
\\
&=& \mathrm{e}^{- \frac{i t \Omega}{2}} \int_{- \infty}^\infty \mathrm{d} x' \sum_{n=0}^\infty \left( \mathrm{e}^{-i t \Omega} \right)^n \langle x | n \rangle \langle n | x' \rangle \Psi(x', 0) \, .
\end{eqnarray}
At this point it is possible to approach this expression by either performing the $x'$ integral first and then the sum (Option 1) or viceversa (Option 2).
Option 1: Integration gives you
\begin{equation}
\int_{- \infty}^\infty \mathrm{d} x' \langle n | x' \rangle \Psi(x', 0) = \left(1 + (-1)^n \right) \frac{\alpha^{1/4}}{\sqrt{2} \left( \alpha + 1 \right)^{1/2}} \frac{(n!)^{1/2}}{\left(\frac{n}{2}\right)!} \left[\frac{i \left( \alpha - 1 \right)^{1/2}}{\sqrt{2}\left( \alpha + 1 \right)^{1/2}}\right]^n \, .
\end{equation}
Then you can use the explicit expression for the eigenfunction $\langle x | n \rangle$ and you will find yourself with sums of the form
\begin{equation}
\sum_{n=0}^{\infty} \frac{k^n}{\left(\frac{n}{2}\right)!} H_n(y) = \frac{4 k^2+2 k y+1}{\left(4 k^2+1\right)^{3/2}} \exp\left[ \frac{4 k^2 y^2}{4 k^2+1} \right] \, .
\end{equation}
I have to thank Mathematica for both results. Putting that together and a bit of algebra gives you the final result.
Option 2: Performing the sum first gives you an intermediate result of greater generality. In this case, we have to deal with the expression
\begin{equation}
\sum_{n=0}^\infty \mathrm{e}^{- \frac{i t \Omega}{2}} \left( \mathrm{e}^{-i t \Omega} \right)^n \langle x | n \rangle \langle n | x' \rangle \,.
\end{equation}
This happens to be the propagator of the quantum harmonic oscillator, also known as the Mehler kernel. This is
\begin{equation}
\langle x | \exp \left[ - i \frac{t H}{ \hbar} \right] | x' \rangle = \sum_{n = 0}^\infty \sum_{m = 0}^\infty \langle x | n \rangle \langle n | \exp \left[ - i \frac{t H}{ \hbar} \right] | m \rangle \langle m | x' \rangle = \sum_{n=0}^\infty \mathrm{e}^{- \frac{i t \Omega}{2}} \left( \mathrm{e}^{-i t \Omega} \right)^n \langle x | n \rangle \langle n | x' \rangle \,.
\end{equation}
Looking for this in references you will find
\begin{equation}
\langle x | \exp \left[ - i \frac{t H}{ \hbar} \right] | x' \rangle = \sqrt{\frac{M \Omega}{\hbar}} \frac{1}{\sqrt{2 \pi i \sin(t \Omega)}} \exp \left\{ \frac{M \Omega }{\hbar } \frac{i}{2 \sin (t \Omega )} \left[ \left(x^2+x'^2\right) \cos (t \Omega )-2 x x' \right] \right\} \, .
\end{equation}
From here, it only remains to perform the $x'$ integral, which looks messy, but in reality it is only a shifted version of a Gaussian integral:
\begin{equation}
\Psi(x,t) = \int_{- \infty}^\infty \mathrm{d} x' \langle x | \exp \left[ - i \frac{t H}{ \hbar} \right] | x' \rangle \Psi(x',0) \, .
\end{equation}
